# A problem on continuous i.i.d random variable

Let $X_1, X_2, X_3, X_4$ be i.i.d continuous random variables with a common distribution function $F$. How to prove that "all the 4! possible orderings of $X_1, \dots, X_4$ are equally likely" without calculating probability of any ordering by integrating the joint density function?

I guess the "exchangablity" property of i.i.d random variable needs to be used. But this property is proved by integrating the joint density function.

Another thing I also want to know the geometric intuitive solution of the problem.

• So far I'm the only person who's up-voted this question. – Michael Hardy Apr 19 '13 at 1:05

The answer is that all orderings are correct "by symmetry". In other words if you were to relabel all of your variables $X_1 => X_2, X_2 => X_3$, etc, you would still have the same probability for the ordering, hence all orderings are equally probable.

Edit: Look at it like this:

Assume that the ordering $X_1, X_2, X_3, X_4$ has probability $p$.

Take another ordering, say $X_4, X_2, X_3, X_1$ and let its probability $= q$.

Now since the random variables are from the same distribution we could, WLOG, have labelled $X_1$ as $X_4$ and $X_4$ as $X_1$.

Hence $P(X_4, X_2, X_3, X_1) = p$ and $p=q$

• I could not understand what is meant by "symmetry" ? How is it possible to tell that "you would still have the same probabilty for the ordering" without calculating the probability ? – RIchard Williams Apr 18 '13 at 17:06
• very nice explanation. – RIchard Williams Apr 19 '13 at 0:24
• @prasenjit thanks I'm glad you found it useful. – phcoding Apr 19 '13 at 11:27

Since our original poster says he can't understand "phcoding"'s answer, I'll try my own rephrasing of it.

The random variables $X_1, X_2, X_3, X_4$ are i.i.d. How do we know that it is just as probable that $$X_4<X_2<X_3<X_1$$ as it is that $$X_1<X_2<X_3<X_4\text{ ?}$$

So let $$W_1=X_4,\qquad W_2=X_2,\qquad W_3=X_3,\qquad W_4=X_1.$$ The question becomes: How do we know it is just as probable that $$W_1<W_2<W_3<W_4$$ as that $$X_1<X_2<X_3<X_4\text{ ?}$$ The answer is that the quadruple $(X_1,X_2,X_3,X_4)$ has the same probability distribution as the quadruple $(W_1,W_2,W_3,W_4)$.

• very nice explanation. – RIchard Williams Apr 19 '13 at 0:26
• @prasenjit : Thank you. – Michael Hardy Apr 19 '13 at 1:03